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Electrolysis - Aluminium extraction

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a. Aluminium is extracted from its ore by an electrolytic process. One extraction plant uses an applied e.m.f. of 5 V, and a current of 225 000 A. One tonne of aluminium is extracted after 14 hours.

i. What is the power consumption of the electrolysis?
ii. What is the effective electrical resistance of the cell?
iii. How much energy, in joules, is used in extracting one tonne of aluminium?

b. The electricity cost is critical for the final price of the aluminium. What cost per kWh of electricity is acceptable if the company wishes to restrict the electricity cost to £300 per tonne or less?

c. Aluminium costs around £1500 per tonne. Suggest three other factors, in addition to the electricity cost, which contribute to this final cost of the raw material.

d. Domestic aluminium foil for food use is 15 × 10-6 m thick. In my local supermarket, a roll of foil measuring 12 m by 0.4 m costs £1.99.

i. Suggest a likely route for the production of the foil, following production of the aluminium in billet form.
ii. Calculate the cost per tonne of the foil as purchased in the supermarket. Account for any difference between this cost and the raw material cost of £1500 per tonne.
The density of aluminium is 2700 kg m-3.
iii. Suggest two reasons, with a brief explanation in each case, why steel is not used to make domestic cooking foil.

e. Explain how recycling aluminium could help to reduce overall costs for the production of the foil. Use information provided on the production of the aluminium foil to inform your answer.

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Solution Summary

This problem relates to the process of metal extraction from it's ore, the production costs of aluminium foil and reduction in cost in recycling process. Includes calculations of power consumption and costs.

Solution Preview

V = 5 V
Electric current,
I = 225000 A
time duration,
t = 14 hr == 14*3600 sec
mass of Al extracted,
m = 1 tonne == 1000 kg

(i). Electric power consumption, P = V*I = 5*225000 = 1125000 W == 1.125 MW

(ii). Electrical resistance, R = V/I = 5/225000 = 2.22*10^(-5) ohm == 22.2 micro-ohm

(iii). Energy used, E = VIt = 5*225000*14*3600 = 5.67*10^10 J

Total energy required/per tonne,
E = 5.67*10^10 J (or W-s) == 5.67*10^7 kW-s == 5.67*10^7/3600 kWh = 15750 kWh

Total electricity ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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